The product of all positive real values of $x$ satisfying the equation $x^{16{\left({\log }_{5}x\right)}^3-68{\log }_{5}x}=5^{-16}$ is ___________ .

The correct answer is 1.

We are given the logarithmic equation:

$x^{16{\left({\log }_{5}x\right)}^3-68{\log }_{5}x}=5^{-16}$

To solve for $x$, let us take the logarithm to the base 5 on both sides of the equation.

${\log }_5\left(x^{16{\left({\log }_{5}x\right)}^3-68{\log }_{5}x}\right)={\log }_5\left(5^{-16}\right)$

Using the power property of logarithms, ${\log }_b\left(a^k\right)=k{\log }_b\left(a\right)$, we can bring the exponents down:

$(16{\left({\log }_{5}x\right)}^3-68{\log }_{5}x)·{\log }_{5}x=-16$

Let us make a substitution to simplify the equation. Let:

$t={\log }_{5}x$

Substituting $t$ into the equation gives:

$(16t^3-68t)·t=-16$

Expanding the left side yields:

$16t^4-68t^2=-16$

Rearranging all terms to one side to form a polynomial equation:

$16t^4-68t^2+16=0$

We can divide the entire equation by 4 to simplify the coefficients:

$4t^4-17t^2+4=0$

This is a quadratic equation in terms of $t^2$. We can factor it by splitting the middle term:

$4t^4-16t^2-t^2+4=0$
$4t^2(t^2-4)-1(t^2-4)=0$
$(4t^2-1)(t^2-4)=0$

This gives us two cases for $t^2$:

1) $4t^2-1=0\Rightarrow t^2=\frac{1}{4}\Rightarrow t=\pm \frac{1}{2}$
2) $t^2-4=0\Rightarrow t^2=4\Rightarrow t=\pm 2$

Thus, the four real roots for $t$ are $t_1=2$, $t_2=-2$, $t_3=\frac{1}{2}$, and $t_4=-\frac{1}{2}$.

Since $t={\log }_{5}x$, we have $x=5^t$. The corresponding values of $x$ are:

$x_1=5^2$
$x_2=5^{-2}$
$x_3=5^{1/2}$
$x_4=5^{-1/2}$

Now, we find the product of all positive real values of $x$:

$P=x_1·x_2·x_3·x_4$
$P=5^2·5^{-2}·5^{1/2}·5^{-1/2}$
$P=5^{2-2+\frac{1}{2}-\frac{1}{2}}$
$P=5^0=1$

Hence, the product of all positive real values of $x$ is 1.